# ADC input circuitry and source impedance

I'm using a maxim ADC in my application from this family: http://datasheets.maxim-ic.com/en/ds/MAX1304-MAX1314.pdf

The devices have relatively low input impedance. If you look at Figure 5 in the above datasheet, on page 19, you can see the equivalent input circuit. My ADC is 0-5V range. My input signal however is in 0-10V range, so i have to reduce it to match the ADC.

If I use a resistive divider at the ADC input, it seems to me that I can not get a proper reading from the ADC due to the interaction between my 2 resistors, and R1 and R2 in figure 5. For example:

|                      Vdiv
|   Vin------Rdivider--------Rdivider---------> GND
|                       |
|                       |
|                       |
|                       R1 (3.33k)
|                       |
|                       |             Csample
|                       |-------/ ------| |-------
|                       |
|                       R2 (5K)
|                       |
|                       |
|                       |
|                       v
|                     0.9V


In the above figure, no matter what Rdivider values I chose, it seems to me that Vdiv will not be Vin/2.

Do I have to use an amplifier in the input for this case?

• Dumb question, but could you get away with using the +/-5V part in the family? One man's 0V is another man's -5V...
– W5VO
Feb 6, 2011 at 7:15
• @W5V0: A good point indeed. Will look into it. Thanks Feb 7, 2011 at 12:16

Yes, from the datasheet:

Due to the analog input resistive divider formed by R1 and R2 in Figure 5, any significant analog input source resistance (R SOURCE) results in gain error. Furthermore, R SOURCE causes distortion due to nonlinear analog input currents. Limit RSOURCE to a maximum of 100Ω.

So practically, this means you need to drive the input with an amplifier. Although, a resistor divider made with a 192 Ω and 200 Ω resistor will meet specs if your signal has a low output impedance.

• Does the 192/200 divider work for low values of Vin? Say, Vin=0.1V? Feb 6, 2011 at 17:03
• Yes, resistors are linear all the way down to 0V and beyond. Feb 6, 2011 at 18:24
• @markrages: Yes, I know, but when you workout the eqns, it doesn't seem like for low voltages, divide by 2 works, due to the 0.9V bias that's connected to R2. So the eqn is: (Vin/Rdivider1) = ((Vin/2)/Rdivider2) + (Vin/2-0.9)/8.3k. The Rdivider values are dependent on the input voltage, due to the bias. Or am I missing something entirely? --regards Feb 7, 2011 at 7:12
• Eek, I didn't notice the 0.9V bias. Feb 7, 2011 at 16:32
• So with a 0V input, and the 192/200 Ω divider, the voltage at Vdiv will be 108 uV. This about 10% of one bit. (5.0 / (1<<12) = 1220 uV). I wouldn't worry about it too much. Feb 7, 2011 at 16:36

I'm sure you've already read this in the datasheet, but it just seems like the easiest solution. They recommend driving the input with a MAX4431 (SOT23-5 package).

• Yes, maybe for the next spin of the pcb. Feb 7, 2011 at 12:17
• Driving it with a op amp as a voltage follower is probably the best idea (just watch out for output oscillation because the ADC is probably quite capacitive and op amps generally don't like this).
– jpc
Mar 21, 2011 at 2:14

How fast is your signal changing and how fast does your sample rate need to be?

If you have a slow (DC) signal, you can add a capacitor between the divider resistors which will supply the sample capacitor as a capacitor's impedance is very low. The capacitor just needs to be big enough to hold a steady voltage during the sample time. This limits your bandwidth significantly though as you now have an RC filter. It also limits your sample rate as sampling too fast can drain the capacitor. It needs time to stabilize between samples, the larger the resistor values the longer this time.

• Thanks for the response. The signal is quite fast, which is why we selected a ~400ksps ADC. Unfortunately, mid-project, the input requirements changed. Feb 7, 2011 at 12:19